When the pressure of a gas doubles, the new volume

Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-1

Irina-Kira [14]

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:

$1 = \%_{16} + \%_{17} + \%_{18}$

$1 = x + 3.2 \cdot 10^{-4} + \%_{18}$

$\%_{18} = 1 - x - 3.2 \cdot 10^{-4}$

Hence, the percent abundance of O-18 is:

$m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18}$

$15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)$

$x = 0.980614 \times 100 = 98.0614 \%$

Hence, the percent abundance of oxygen-18 is:

$\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%$

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!

8 0

3 years ago

Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation

sleet_krkn [62]

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

7 0

3 years ago

What is the final step in the scientific method

kherson [118]

The final step in the scientific method is the conclusion. The conclusion will either clearly support the hypothesis or it will not. If the results support the hypothesis a conclusion can be written

8 0

3 years ago

Read 2 more answers

The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van

Goshia [24]

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: Δ $T_{b}$ =i $K_{b}$ .m, where Δ $T_{b}$ = elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

$K_{b}$ = Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i= $\frac{2.5}{0.512 X 1.83}$

i=2.66= 2.7 (approx.)

6 0

3 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

8 0

3 years ago